# Prove an integral inequality $|\int\limits_0^1f(x)dx|\leq\frac{1-a+b}{4}M$ [closed]

Let $f$ be a differentiable function on $[0,1]$ and $a,b\in(0,1)$ such that $a<b$, $\int\limits_0^af(x)dx=\int\limits_b^1f(x)dx=0$. Show that: $$\left|\int_0^{1} f(x)\,dx\,\right|\leq\frac{1-a+b}{4}\,M$$ where $M=\sup\limits_{x\in[0,1]}|f'(x)|$.

## closed as off-topic by Ben Sheller, Matthew Conroy, man and laptop, GNUSupporter 8964民主女神 地下教會, kimchi loverFeb 8 '18 at 0:27

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• What are your ideas on this problem? – Dan Mar 2 '15 at 12:31
• So in fact you're assuming as well the derivative is bounded on the unit interval? – Timbuc Mar 2 '15 at 12:54
• @ Matthias - you're missing the condition on the integrals between $0$ and $a$, and $b$ and $1$. – Baron Mingus Mar 2 '15 at 13:00
• @ Matthias $\int_0^a x \mathrm{d}x \neq 0$ for $a > 0$. – Baron Mingus Mar 2 '15 at 13:06
• Upps I missed the last equality. Thx. – Matthias Mar 2 '15 at 13:07

## 1 Answer

Since $f$ is continuous, it must have zeroes in both intervals $[0,a]$ and $[b,1]$. By the Mean value theorem, we deduce that $$|f(a)| \leq Ma,\qquad |f(b)| \leq M(1-b).$$

By the Mean value theorem again, there exists $c \in (a,b)$ such that $$\frac{1}{b-a}\left[\int_a^bf(x)dx - \frac{1}{2}(b-a) [f(a)+f(b)]\right] = -(c-\frac{a+b}{2})f'(c).$$ Therefore, $$\left|\int_a^b f(x)dx\right| \leq \frac{b-a}{2} \left(|f(a)| + |f(b)| + (b-a)M\right) \leq \frac{b-a}{2}M.$$ The conclusion follows since $b-a < 1$ implies $\frac{b-a}{2} < \frac{b-a + 1}{4}$ and $$\int_0^1 f(x)dx = \int_a^b f(x)dx.$$

• It may help to note that in the second step he applied the mean value theorem to the function $g(x)=(\frac{a+b}{2}-x)f(x)+\int_0^x f(t) dt$. – J.R. Mar 2 '15 at 14:29